Do we not need exception specification for the main method in a Java program. For example, the following code works exactly the same without specifying "throws Xcept" for the main method.
class Xcept extends Exception {
public Xcept(){
}
public Xcept(String msg){
super(msg);
}
}
public class MyException {
public void f() throws Xcept {
System.out.println("Exception from f()");
throw new Xcept("Simple Exception");
}
public static void main(String[] args) throws Xcept {
MyException sed = new MyException();
try {
sed.f();
} catch(Xcept e) {
e.printStackTrace();
}
finally {
System.out.println("Reached here");
}
}
}
I read that java enforces this, but I don't get a compile time error if I exclude this specification for the main method.
That's because Xcept will never be thrown out of your main method, as you actually catch it there... The sed.f() call may result in an Xcept being thrown, but it's caught and handled.
Related
If I add the throws keyword in the method signature and handle it in my main() inside my TestClass{}, is it a good approach or should I handle the exception in the method in which it occurs fooBar(), which approach is recommended or what are the trade-offs between the two. Sorry this question might seem a little weird, just a newbie in Java.
public class Foo {
//private fields
public Foo() {
//ctor
}
//accessors and mutators
//method that throws exception
public void fooBar() throws FooBarException {
throw new FooBarException();
}
}
Generally, the scope of try block should be as small as possible.
public void fooBar(){
try{
//Lines that can through error
}
catch (FileNotFoundException e) {
//Handle exception Here
//e.printStackTrace();
/*
other code that will not throw any error
*/
}
or if the whole block may throw the error then
public void fooBar() throws FooBarException {
}
//Handle in the exception where you calling the method
I don't think I understand the try-catch block and throws really.
public class TestException {
public static void main(String[] args) {
new TestException().tt();
}
public void tt() {
try {
throw new RuntimeException();
}catch (Exception e) {
throw e;
}
}
}
When in Eclipse, there is an error hint about 'Unhandled exception type xxx', and if you run this, you will get an
Exception in thread "main" java.lang.Error: Unresolved compilation problem:
Unhandled exception type Exception
But in Idea, there's no errors. It runs and throws the exception correctly.
In my opnion, the 'e' was not declared as a RuntimeException(althrough it is an RuntimeException), so the tt() method must be declared with throws. But actually it's not. Why?
This should answer your question:
public class TestException {
public static void main(String[] args) {
new TestException().tt();
}
public void tt() {
try {
throw new RuntimeException();
}catch (Exception e) {
throw new RuntimeException(e);
}
}
}
If you use throws, you tell those who use your function, "My function may throw exceptions. You have to handle that."
You should get difference of throws and throw in this sentence.
If you use try-catch, you handle that exception.
1) You should add throws keyword like below
public class TestException {
public static void main(String[] args) {
new TestException().tt();
}
public void tt() **throws Exception** {
try {
throw new RuntimeException();
}catch (Exception e) {
throw e;
}
}
}
2) Handle exception where you use function
public class TestException {
public static void main(String[] args) {
try{
new TestException().tt();
}catch(Exception e){
System.out.println("Error handled");
}
}
public void tt() throws Exception {
try {
throw new RuntimeException();
}catch (Exception e) {
throw e;
}
}
}
In general if you catch an exception you handle it. Like this
public class TestException {
public static void main(String[] args) {
new TestException().tt();
}
public void tt() {
try {
throw new RuntimeException();
}catch (Exception e) {
System.out.println("Error caught! ")
}
}
}
or
public class TestException {
public static void main(String[] args) {
try {
new TestException().tt();
}catch(Exception e){
System.out.println("Error caught! ")
}
}
public void tt() throws RuntimeException {
throw new RuntimeException();
}
}
You can also throw other's exception
public class TestException {
public static void main(String[] args) {
try{
new TestException().a();
}catch(Exception e){
System.out.println("Error handled");
}
}
public void a() throws Exception {
b();
}
public void b() throws Exception {
c();
}
public void c() throws Exception {
throw new RuntimeException();
}
}
I think that you want to look into 'Checked Exceptions' and 'Unchecked Exceptions'.
Only Checked Exceptions need to be declared in a methods signature, and RuntimeException is an unchecked exception (though you can declare it if you like - it just isn't necessary to compile).
https://docs.oracle.com/javase/7/docs/api/java/lang/Exception.html
The API for java Exception says:
"The class Exception and any subclasses that are not also subclasses of RuntimeException are checked exceptions. Checked exceptions need to be declared in a method or constructor's throws clause if they can be thrown by the execution of the method or constructor and propagate outside the method or constructor boundary"
https://docs.oracle.com/javase/7/docs/api/java/lang/RuntimeException.html
The API for java RuntimeException says:
"RuntimeException and its subclasses are unchecked exceptions. Unchecked exceptions do not need to be declared in a method or constructor's throws clause if they can be thrown by the execution of the method or constructor and propagate outside the method or constructor boundary."
In my opnion, the 'e' was not declared as a RuntimeException(althrough it is an RuntimeException), so the tt() method must be declared with throws. But actually it's not. Why?
Let's consider what we know:
When using rethrow syntax, the existing exception object (e) is rethrown.
e is an object of class Exception, or one of its subtypes.
RuntimeException is a subtype of exception, and is not compiled time checked, so it's possible the re-thrown object is a non compile time checked object.
The compiler cannot see a place where the code definitely, or even possibly throws a compile checked exception, and so it makes sense that it does not force those semantics.
For example, if you change your catch to an IOException, the compiler will not allow you to do that without a line in the try which could possibly lead to an IOException.
If you added such a line, then the compiler would recognize that the throw would rethrow a compile time checked exception, and make you catch it again, or mark the function with the appropriate throws clause.
As for eclipse, your code compiles OK in mine with my JDK.
I want to throw an exception (any type) in Java, but the restriction is that i can't add " throws Exception " to my main method. So i tried this:
import java.io.IOException;
class Util
{
#SuppressWarnings("unchecked")
private static <T extends Throwable> void throwException(Throwable exception, Object dummy) throws T
{
throw (T) exception;
}
public static void throwException(Throwable exception)
{
Util.<RuntimeException>throwException(exception, null);
}
}
public class Test
{
public static void met() {
Util.throwException(new IOException("This is an exception!"));
}
public static void main(String[] args)
{
System.out.println("->main");
try {
Test.met();
} catch (Exception e) {
System.out.println(e.getMessage());
}
}
}
This code works, but when i am trying to catch an "IOException", for examle, in try-catch block, it doesnt compile. The compiler tells me that IOException is never thrown. It works only for exceptions that extend RuntimeException. Is there a way to solve this?
Added:
import java.io.IOException;
class Util
{
#SuppressWarnings("unchecked")
private static <T extends Throwable> void throwException(Throwable exception, Object dummy) throws T
{
throw (T) exception;
}
public static void throwException(Throwable exception)
{
Util.<RuntimeException>throwException(exception, null);
}
}
public class Test
{
public static void met() { // this method's signature can't be changed
Util.throwException(new IOException("This is an exception!"));
}
public static void main(String[] args)
{
System.out.println("->main");
try {
Test.met();
} catch (IOException e) { // can't be changed and it does not compile right now
System.out.println(e.getMessage());
}
}
}
The simple answer: you can't.
The more complex answer: you can't, and you really shouldn't look to do this. The main reason being, if your code can catch exceptions that it's not advertised to, then that leads to inconsistency and bugs.
Above all, that code block isn't meant to catch anything other than an IOException; that is, the code is only meant to recover on something going haywire with IO. If I were to try and catch anything else, then that would imply that the code knows how to recover from that scenario, which is very much not the case.
As an aside, any children of IOException will be caught by that block, so you don't have to worry about catching FileNotFoundExecption, since that will handle it.
This is awful coding, and I feel dirty just writing it...
Instead of catch-ing the IOException directly, you can check that the caught Exception is an IOException.
public class Test
{
public static void main(String[] args)
{
System.out.println("->main");
try {
Test.met();
} catch (Exception e) {
if (e instanceof IOException) {
System.out.println(e.getMessage());
}
}
}
}
I know that if we have a normal code without try and catch statements,then if an exception occurs,then the default exception handler of JVM handles that exception.
I have a code...
public class St
{
public static void main(String args[])
{
try
{
int y=23/0;
}
catch(Exception e)
{
System.out.println("Division by zero");
}
}
}
As far as I know, in this code exception occurs at line 7,an object of class Exception is thrown and that's why we have taken as argument an object of class Exception in order to catch the exception.Am I right upto now????
But why this code shows a compile time error...
public class St
{
public static void main(String args[])
{
Exception e=new Exception();
try
{
int y=23/0;
}
catch(e)
{
System.out.println("Division by zero");
}
}
}
In this I have created an object reference e of class Exception,and that I have taken as argument in catch.But its not running,giving error at compile time.Can someone explain why???
That's just not how the catch block works. It requires an ExceptionType argument and then a name to reference the exception once it's been caught. It doesn't take an object as an argument, but the name of a class that inherits from 'Throwable'.
Blow is a example copied from Think in Java 4 edition
public class Rethrowing {
public static void f() throws Exception {
System.out.println("originating the exception in f()");
throw new Exception("thrown from f()");
}
public static void h() throws Exception {
try {
f();
} catch (Exception e) {
System.out.println("Inside h(),e.printStackTrace()");
e.printStackTrace(System.out); //first print line throw (Exception) e.fillInStackTrace();
}
}
public static void main(String[] args) {
try {
h();
} catch (Exception e) {
System.out.println("main: printStackTrace()");
e.printStackTrace(System.out);
}
}
}
Output:
originating the exception in f()
Inside h(),e.printStackTrace()
java.lang.Exception: thrown from f()
at Rethrowing.f(Rethrowing.java:20)
at Rethrowing.h(Rethrowing.java:25)
at Rethrowing.main(Rethrowing.java:35)
main: printStackTrace()
java.lang.Exception: thrown from f()
at Rethrowing.f(Rethrowing.java:20)
at Rethrowing.h(Rethrowing.java:25)
at Rethrowing.main(Rethrowing.java:35)
When comment //first print line
Output:
originating the exception in f()
Inside h(),e.printStackTrace()
main: printStackTrace()
java.lang.Exception: thrown from f()
at Rethrowing.h(Rethrowing.java:29)
at Rethrowing.main(Rethrowing.java:35)
My question is why i first invoke e.printStackTrace(printOut out ) method before the fillInStackTrace mehod, and then the fillInStackTrace seems not available. Any one can do me a faver, thanks in advance.
user917879, When you call e.fillInStackTrace(); it resets the StackTrace. So, to print the current StackTrace - before it is get reset - you need to invoke the e.printStackTrace(printOut out ) first.